Question

44.At 25

°

C, the equilibrium partial pressures of

NO₂

and

N₂O₄

are

0.150 atm

and

0.200 atm,

respectively. If the volume is increased by 2.20 fold at constant temperature, calculate the partial pressures of the gases when a new equilibrium is established.

PNO2 =  atm

PN2O4 =  atm

45. At 25

°

C, the equilibrium partial pressures of

NO₂

and

N₂O₄

are

0.150 atm

and

0.200 atm,

respectively. If the volume is increased by 2.20 fold at constant temperature, calculate the partial pressures of the gases when a new equilibrium is established.

PNO2 =  atm

PN2O4 =  atm

Answer 1

N₂O₄                  2NO₂

Given, P(N₂O₄) = 0.2 atm

P(NO₂) = 0.15 atm

Now,

Kp = [(PNO₂)²]/[PN₂O₄] = [(0.15)2/(0.2)] = 0.1125

If the volume is increased by 2.20 fold at constant temperature, therewill be a proportional decrease in the pressure

P(N₂O₄) = 0.2/2.2 = 0.09 atm

P(NO₂) = 0.15/2.2 = 0.068 atm

Q = (0.068)²/(0.09) = 0.051

Here, Q < Kp. Therefore, the reaction will proceed in the forward direction

                N₂O₄                   2NO₂

Initial       0.09 atm              0.068 atm

Final        (0.09 - x)              (0.068 + 2x)

Kp = 0.1125 = (0.068 + 2x)²/(0.09 - x)

0.1125 x (0.09 - x) = (0.068 + 2x)²

0.010125 – 0.1125x = 0.004624 + 0.272 x + 4x²

4x² + 0.3845x -0.0055 = 0

On solving, x = 0.0126

Therefore, P(N₂O₄) = (0.09 -0.0126) = 0.0774

P(N₂O₄) = 0.0774

P(NO₂) = (0.068 + 2 x 0.0126) = 0.0932

P(NO₂) = 0.0932